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andrew-mc [135]
1 year ago
7

a sample of a metal, heated in air, gains mass dm by oxidising at a rate that declines with time t in a parabolic way, meaning t

hat dm2 ¼ kpt where kp is a constant. why does the rate fall off in this way? what does it imply about the nature and utility of the oxide?
Chemistry
1 answer:
velikii [3]1 year ago
5 0

Oxidation originates at the surface of the metal and then it slowly penetrates through the surface.

<em>As at the beginning exposed surface area is more therefore rate of oxidation is also more. After that, the process needed to occur inside the exposed surface of the metal, and the rate of diffusion decreases, so the rate of oxidation decreases.</em>

<em />

<em>It represents oxide layers that reduce the diffusion rate, therefore it is used for producing isolation shields for metals. Apart from that for covering a metal from the external environment or masking oxide layers are produced on metal surfaces.</em>

<em />

<em />

Mass is a quantitative measure of inertia, an essential property of all dependents. It's far, in effect, the resistance that a frame of count number gives to a change in its speed or position upon the application of pressure. The more the mass of a frame, the smaller the alternate produced with the aid of a carried-out force.

Learn more about Mass here:-brainly.com/question/28021242

#SPJ4

<em />

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Manganese forms several oxides when combined with oxygen. One of the oxides (Oxide 1) contains 63.2% of Mn and another oxide (Ox
Nina [5.8K]

Explanation:

Defining law of definite proportions, it states that when two elements form more than one compound, the ratios of the masses of the second element which combine with a fixed mass of the first element will always be ratios of small whole numbers.

A. One of the oxides (Oxide 1) contains 63.2% of Mn.

Mass of the oxide = 100g

Mass of Mn = 63.2 g

Mass of O = 100 - 63.2

= 36.8 g

Ratio of Mn to O = 63.2/36.8

= 1.72

Another oxide (Oxide 2) contains 77.5% Mn.

Mass of oxide = 100 g

Mass of Mn = 77.5 g

Mass of O = 100 - 77.5

= 22.5 g

Ratio of Mn to O = 77.5/22.5

= 3.44

Therefore, the ratio of the masses of Mn and O in Oxide 1 and Oxide 2 is in the ratio 1.72 : 3.44, which is also 1 : 2. So the law of multiple proportions is obeyed.

B.

Oxide 1

Mass of Mn per 1 g of O = mass of Mn/mass of O

= 77.5/22.5

= 3.44 g/g of Oxygen.

Oxide 2

Mass of Mn per 1 g of O = mass of Mn/mass of O

= 77.5/22.5

= 3.44 g/g of Oxygen.

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3 years ago
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What is the lightest particle in the following 1)electron 2)proton 3)neutron 4)photon
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Answer:

1) Electron

Explanation:

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5 0
2 years ago
Read 2 more answers
Hello, everyone!
marusya05 [52]

Answer: 27.09 ppm and 0.003 %.

First, <u>for air pollutants, ppm refers to parts of steam or gas per million parts of contaminated air, which can be expressed as cm³ / m³. </u>Therefore, we must find the volume of CO that represents 35 mg of this gas at a temperature of -30 ° C and a pressure of 0.92 atm.

Note: we consider 35 mg since this is the acceptable hourly average concentration of CO per cubic meter m³ of contaminated air established in the "National Ambient Air Quality Objectives". The volume of these 35 mg of gas will change according to the atmospheric conditions in which they are.

So, according to the <em>law of ideal gases,</em>  

PV = nRT

where P, V, n and T are the pressure, volume, moles and temperature of the gas in question while R is the constant gas (0.082057 atm L / mol K)

The moles of CO will be,

n = 35 mg x \frac{1 g}{1000 mg} x \frac{1 mol}{28.01 g}

→ n = 0.00125 mol

We clear V from the equation and substitute P = 0.92 atm and

T = -30 ° C + 273.15 K = 243.15 K

V =  \frac{0.00125 mol x 0.082057 \frac{atm L}{mol K}  x 243 K}{0.92 atm}

→ V = 0.0271 L

As 1000 cm³ = 1 L then,

V = 0.0271 L x \frac{1000 cm^{3} }{1 L} = 27.09 cm³

<u>Then the acceptable concentration </u><u>c</u><u> of CO in ppm is,</u>

c = 27 cm³ / m³ = 27 ppm

<u>To express this concentration in percent by volume </u>we must consider that 1 000 000 cm³ = 1 m³ to convert 27.09 cm³ in m³ and multiply the result by 100%:

c = 27.09 \frac{cm^{3} }{m^{3} } x \frac{1 m^{3} }{1 000 000 cm^{3} } x 100%

c = 0.003 %

So, <u>the acceptable concentration of CO if the temperature is -30 °C and pressure is 0.92 atm in ppm and as a percent by volume is </u>27.09 ppm and 0.003 %.

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3 years ago
PLEASE HELP!!!!! Noble gases are often described as being very stable. Give two points to explain what this means.
Leya [2.2K]

Answer:

Noble Gases are very stable because...

1) Noble Gases have a full octet

2) Low chemical reactivity

Explanation:

This means that is has 8 valence electrons so it won't need any more, so it is less likely to react with other substances.

6 0
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